A physics student stands at the top of a hill that has an elevation of 56 meters. he throws a rock and it goes up into the air and then falls back past him and lands on the ground below. The path of the rock can be modeled by the equation y=0.04x^2+1.3x+56 where x is the horizontal distance, in meters, from the starting point on the top of the hill and y is the height, in meters, of the rock above the ground.

Question

luigi0210 · Answer

What exactly is the question?

anonymous · Answer

How far horizontally  from its starting point will the rock land?

amistre64 · Answer

hills come in all shapes ... an elevation tells us how high it is but not how wide it is .....

amistre64 · Answer

|dw:1397137401644:dw|

anonymous · Answer

Yea I know. But that's how it was written on my homework page. It doesn't make any sense to me..

amistre64 · Answer

find the roots of the quadratic equation and assume they model the ground level .... without  knowing the slope of the hillside it would be a bit more complicared

amistre64 · Answer

and since the leading coefficient of your quadratic is positive, the rock seems to want to rise up to the moon instead of fall to the ground

anonymous · Answer

How do you find the roots? I wasn't in school when they showed how to do it.

amistre64 · Answer

one of the simplest ways is to use the quadratic formula:

given:  ax^2 + bx + c = 0

x = -b +- sqrt(b^2-4ac)
      ------------------
                   2a

anonymous · Answer

Okay thank you

amistre64 · Answer

your welcome ... if we want to make the setup more general then we would determine where the line y=mx+56 intersects the rocks path and assess the distances that way

anonymous · Answer

Oohhh OK now I kinda get it thank you! You are very smart!

amistre64 · Answer

i have my moments :)